The South African Mathematical Olympiad Third Round

Note first that there is no calm representation with 2 terms: if either $a_1$ or $a_2$ is 1 or $a_1=a_2=\frac{1}{2}$, then the sum is greater than $\frac{17}{20}$. Otherwise, the sum is at most $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}<\frac{17}{20}$, thus too small.

On the other hand, there is a representation with 3 terms, namely $$\frac{17}{20}=\frac{1}{2}+\frac{1}{4}+\frac{1}{10},$$ showing that the smallest possible value of $k$ is 3.

Now we show that there are infinitely many calm representations: take $a_1=5\cdot 2^{4n+1}$ and consider the difference $$\frac{17}{20}-\frac{1}{5\cdot 2^{4n+1}}=\frac{17\cdot 2^{4n-1}-1}{5\cdot 2^{4n+1}}.$$ Since $2^4=16\equiv 1\mathrm{mod}5$, the numerator is $17\cdot 2^{4n-1}-1\equiv 17\cdot 8-1=135\equiv 0\mathrm{mod}5$. Thus the factor 5 cancels, and we have $$\frac{17}{20}-\frac{1}{5\cdot 2^{4n+1}}=\frac{A}{2^{4n+1}}$$ for some positive integer $A<2^{4n+1}$. Since $A$ has a binary representation as $A=2^{b_1}+2^{b_2}+\cdots +2^{b_r}$ with distinct nonnegative integers $b_1,b_2,\dots ,b_r$, we get $$\frac{17}{20}=\frac{1}{5\cdot 2^{4n+1}}+\frac{1}{2^{4n+1-b_1}}+\frac{1}{2^{4n+1-b_2}}+\cdots +\frac{1}{2^{4n+1-b_r}},$$ which is a calm representation for every $n$. This completes the proof.